Strain

In an earlier blog, the one with the example of stretched rubber tubing, we included a description of the concept of strain.  Now, we try to make it more plausible, more precise, and to include shearing, as well as normal strain.

We'll start with the one-dimensional case again.  We choose a fixed coordinate system, and consider a surface somewhere inside a material body, at a point x = (x,y,z), where the plane tangent to the surface is the y-z plane.  That is, the normal vector n(x) = (1,0,0). 


Under the action of forces applied to the surface of the body, let the displacement of the point x in the x direction be u(x).  Next, consider a point at a distance a further along the x-axis,  at x+a.  What is the displacement u (x+a)?  We can find out by expanding u(x+a) in a Taylor series (Wikipedia has a good article on the latter).  The expansion is

 

where u' is the first derivative of u with respect to x, u" is the second derivative, and so on.  If we take a to be small enough, then we can drop all the terms with higher powers of a, keeping only the first two.  Then, for a sufficiently small,  we have [u(x+a) - u(x)]/a = du/dx.  This is called the normal strain in the x-direction.  (It's also simply the definition of the derivative of u(x)). Later we will use the observation that [u(x+a)-u(x)]/a  is equal to the fractional change in length of the line segment (originally of length a) joining the points x and x+a.


The material is said to be in tension when the normal strain is positive.  Note that the displacement is then larger at x+a than it is at x.  So as we noted with the rubber tubing, the displacement must increase in the direction of increasing x.  Furthermore, we noted that for materials that follow Hooke's law, the normal displacement is a linear function of x.  


In the previous blog, we found that for a body in mechanical equilibrium, the divergence of the stress tensor is zero, which means that the stress is constant (i.e., doesn't depend on x) for this one-dimensional example.  Therefore each point in the material along x must be in tension.  


For example, when you pull on the rubber tubing at one end (holding the left end fixed in space), the material at the point x moves to the right.  Therefore the material at the point x+a must also move to the right - otherwise the material at x+a must also move to the right, since it is under the same stress.  So the material at x+a must move the same distance to the right as the material at x, plus an additional distance in order to be in tension. For a material following Hooke's law, the fractional  (i.e., percentage) change in position of each point is constant.  


The purpose of all this discussion is to make Hooke's law seem plausible.  That is, if stress and strain are both constant in a one-dimensional linearly elastic material, it makes sense that they should be proportional to each other, with the coefficient of proportionality being a property of the material.  


As an aside, note that we are assuming the material is isotropic, i.e. that its properties are the same in all directions.  For example if the elasticity varied so that, for example, the rubber was "stiffer" at x+a than at x, it would be possible for the displacement at x+a to be less than at x.  


However, things get more complicated in the three-dimensional case, as we'll see in the next blog.  Since stress is a tensor, it seems reasonable that strain is also, and we need to find a linear relationship between these two tensors.